# How to prove $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^{\infty}{\frac{1}{k!}}=:e$ [duplicate]

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I have already proven that $$\left(1+\frac 1{n}\right)^n\leq \sum_{k=0}^{n}{\frac{1}{k!}}$$ . Does that help me in any way? I am stumped...

Any suggestions/hints?

## marked as duplicate by Jyrki Lahtonen, Xander Henderson, Shailesh, YuiTo Cheng, LeucippusJun 8 at 0:43

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## 3 Answers

If $$n,N\in\mathbb N$$, with $$n\geqslant N$$, then\begin{align}\left(1+\frac1n\right)^n&=\sum_{k=0}^n\binom nk\frac1{n^k}\\&\geqslant\sum_{k=0}^N\binom nk\frac1{n^k}\\&=1+1+\frac{n-1}{2n}+\frac{(n-1)(n-2)}{3!n^2}+\cdots+\frac{(n-1)(n-2)\ldots\bigl(n-(N-1)\bigr)}{N!n^{N-1}}\\&=1+1+\frac1{2!}\left(1-\frac1n\right)+\frac1{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots\\&\phantom{=\ }+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{N-1}n\right).\end{align}Therefore,\begin{align}\left(1+\frac1n\right)^n&\geqslant\lim_{n\to\infty}1+1+\cdots+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\frac1{2!}\left(1-\frac1n\right)\cdots\left(1-\frac{N-1}n\right)\\&=1+1+\frac1{2!}+\cdots+\frac1{N!}\end{align}and so$$\lim_{n\to\infty}\left(1+\frac1n\right)^n\geqslant1+1+\frac1{2!}+\cdots+\frac1{N!}.$$Since this takes place for every $$N\in\mathbb N$$,$$\lim_{n\to\infty}\left(1+\frac1n\right)^n\geqslant\sum_{k=0}^\infty\frac1{k!}.$$

• \begin{align}\left(1+\frac1n\right)^n&\geqslant\lim_{n\to\infty}1+1+\cdots+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\frac1{2!}\left(1-\frac1n\right)\cdots\left(1-\frac{N-1}n\right)\\&=1+1+\frac1{2!}+\cdots+\frac1{N!}\end{align} How do you get that Inequality? – ParabolicAlcoholic Jun 8 at 11:04
• No. I did no use that. – José Carlos Santos Jun 8 at 11:09
• But how do you know that? – ParabolicAlcoholic Jun 8 at 11:27
• How do I know what? Are you talking about the first inequality of my proof? – José Carlos Santos Jun 8 at 11:36
• Ok, sorry, I misinterpreted somthing. It's clear now. – ParabolicAlcoholic Jun 8 at 11:45

I don't know if this is what you're looking for but:

let exp(x) be the function that satisfies the D.E.: $$\frac{d}{dx}\exp(x) = \exp(x),\qquad \exp(0) = 1$$ hence if we do a Taylor expansion of this function we get: $$\exp(x) = \sum_{k=0}^\infty{\frac{x}{k!}}$$ using the limit definition of the derivative with a constant to the power x: $$\frac{d}{dx}\left(k^x\right) = \lim \limits_{h \to 0}\left(\frac{k^{x+h} - k^{x}}{h}\right) = \lim \limits_{h \to 0}\left(k^{x}\left(\frac{k^{h} - 1}{h}\right)\right) = T\cdot k^{x}$$ where $$T = \lim \limits_{h \to 0}\left(\frac{k^{h} - 1}{h}\right)$$ is constant with respect to x. Let e be the number such that $$T = 1$$, trivially, $$e^x = \exp(x)$$. rearranging our formula for T: $$\lim \limits_{h \to 0}\left[\frac{e^{h} - 1}{h}\right] = 1 \Rightarrow \lim \limits_{h \to 0}\left[e =\left(1+h\right)^\frac{1}{h}\right]$$ letting $$u = \frac{1}{h}$$: $$e =\lim \limits_{u \to \infty}\left[\left(1+\frac{1}{u}\right)^u\right]$$ hence: $$\lim \limits_{x \to \infty}\left[\left(1+\frac{1}{x}\right)^x\right] = e = \exp(1) = \sum_{k=0}^\infty{\frac{1}{k!}}$$

For all $$n\in\mathbb N$$ $$\frac n{n+1}\le\ln\left(1+\frac1n\right)^n\le1.$$ 